continuous derivative implies bounded variation

Theorem.  If the real function $f$ has continuous derivative on the interval  $[a,b]$,  then on this interval,

• $f$ is of bounded variation,

• $f$ can be expressed as difference of two continuously differentiable monotonic functions.

Proof.  $1^{\underset{¯}{o}}$. The continuous function $|f^{\prime }|$ has its greatest value $M$ on the closed interval  $[a,b]$,  i.e.

$$|f^{\prime }(x)|\leqq M\mathit{\hspace{1em}}\forall x\in [a,b].$$

Let $D$ be an arbitrary partition of  $[a,b]$,  with the points

Consider $f$ on a subinterval  $[x_{i-1},x_i]$.  By the mean-value theorem, there exists on this subinterval a point ${\xi }_i$ such that  $f(x_i)-f(x_{i-1})=f^{\prime }({\xi }_i)(x_i-x_{i-1})$.  Then we get

$$S_D:=\sum _{i=1}^n|f(x_i)-f(x_{i-1})|=\sum _{i=1}^n|f^{\prime }({\xi }_i)|(x_i-x_{i-1})\leqq M\sum _{i=1}^n(x_i-x_{i-1})=M(b-a).$$

Thus the total variation satisfies

whence $f$ is of bounded variation on the interval  $[a,b]$.

$2^{\underset{¯}{o}}$. Define the functions $G$ and $H$ by setting

$$G:=\frac{|f^{\prime }|+f^{\prime }}{2},H:=\frac{|f^{\prime }|-f^{\prime }}{2}.$$

We see that these are non-negative and that  $f^{\prime }=G-H$.  Define then the functions $g$ and $h$ on  $[a,b]$  by

$$g(x):=f(a)+{\int }_a^{x}G(t)𝑑t,h(x):={\int }_a^{x}H(t)𝑑t.$$

Because $G$ and $H$ are non-negative, the functions $g$ and $h$ are monotonically nondecreasing.  We have also

$$(g-h)(x)=f(a)+{\int }_a^x(G(t)-H(t))𝑑t=f(a)+{\int }_a^x{f}^{\prime }(t)𝑑t=f(x),$$

whence  $f=g-h$.  Since $G$ and $H$ are by their definitions continuous, the monotonic functions $g$ and $h$ have continuous derivatives  $g^{\prime }=G$,  $h^{\prime }=H$.  So $g$ and $h$ fulfil the requirements of the theorem.

Remark.  It may be proved that each function of bounded variation is difference of two bounded monotonically increasing functions.